Question:
If the height of a transmitting antenna is \(45\,\mathrm{m}\) and the height of the receiving antenna is \(H\), then the maximum line-of-sight distance between the two antennas is \(1\%\) of the radius of the earth. If the radius of the earth is \(6400\,\mathrm{km}\), then the value of \(H\) is
If the height of a transmitting antenna is \(45\,\mathrm{m}\) and the height of the receiving antenna is \(H\), then the maximum line-of-sight distance between the two antennas is \(1\%\) of the radius of the earth. If the radius of the earth is \(6400\,\mathrm{km}\), then the value of \(H\) is
Show Hint
The maximum line-of-sight distance between two antennas is
\[
\boxed{
d=\sqrt{2Rh_1}+\sqrt{2Rh_2}
}
\]
where all distances are in the same units.
Updated On: Jul 15, 2026
- \(125\,\mathrm{m}\)
- \(75\,\mathrm{m}\)
- \(90\,\mathrm{m}\)
- \(150\,\mathrm{m}\)
Show Solution
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The Correct Option is A
Solution and Explanation
Step 1: Calculate the line-of-sight distance.
The maximum line-of-sight distance is
\[
d
=
\sqrt{2Rh_1}
+
\sqrt{2Rh_2}.
\]
Given,
\[
d
=
1\%\ \text{of Earth's radius}
=
0.01\times6400
=
64\,\mathrm{km},
\]
\[
R
=
6400\,\mathrm{km},
\]
\[
h_1
=
45\,\mathrm{m}
=
0.045\,\mathrm{km}.
\]
Step 2: Substitute the values. \[ 64 = \sqrt{2\times6400\times0.045} + \sqrt{2\times6400\times H'}, \] where \(H'\) is in km. Since \[ \sqrt{576}=24, \] \[ 64=24+\sqrt{12800H'}. \] Thus, \[ \sqrt{12800H'} = 40, \] \[ 12800H' = 1600, \] \[ H' = 0.125\,\mathrm{km} = 125\,\mathrm{m}. \] Hence, \[ \boxed{125\,\mathrm{m}} \] Therefore, \[ \boxed{(A)} \] is the correct answer.
Step 2: Substitute the values. \[ 64 = \sqrt{2\times6400\times0.045} + \sqrt{2\times6400\times H'}, \] where \(H'\) is in km. Since \[ \sqrt{576}=24, \] \[ 64=24+\sqrt{12800H'}. \] Thus, \[ \sqrt{12800H'} = 40, \] \[ 12800H' = 1600, \] \[ H' = 0.125\,\mathrm{km} = 125\,\mathrm{m}. \] Hence, \[ \boxed{125\,\mathrm{m}} \] Therefore, \[ \boxed{(A)} \] is the correct answer.
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